I was reading the CSS ( Steane Code) from the Nielsen & Chuang book. It asked in Ex. 10.27 to prove that: suppose $C_1$ and $C_2$ are $[n,k_1]$ and $[n,k_2]$classical linear codes such that $C_2\subset C_1$ and $C_1$ and $C_2^\perp$ both correct $t$ errors. Codes defined by $$|x+C_2\rangle\equiv \dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}|x+y+v\rangle $$ and parametrized by $u$ and $v$ are equivalent to $\mathrm{CSS}(C_1, C_2)$ in the sense that they have the same error-correcting properties.
My attempt for this was let the corrupted state be for the bit flip case: $$\dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}|x+y+v+e_1\rangle$$ now proceeding on the lines of the code $\mathrm{CSS}(C_1,C_2)$, I apply the Parity matrix $H_1$ for $C_1$, on the ancilla to obtain $$ \dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}|x+y+v+e_1\rangle|H_1e_1\rangle$$ where $H_1(x+y+v)=0$. so i get the position of the flipped qubit by inspecting the position where $1$ occurs.
Now for the phase flip case here is my try $$\dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).e_2}|x+y+v\rangle $$ now I apply the Hadamard gate on the qubit to obtain $$\dfrac{1}{\sqrt{|C_2|2^n}}\sum_z\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).e_2}(-1)^{(x+y+v).z}|z\rangle $$$$= \dfrac{1}{\sqrt{|C_2|}2^n}\sum_z\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).(e_2+z)}|z\rangle$$ Now let $e_2+z=z'$, we get $$ \dfrac{1}{\sqrt{|C_2|2^n}}\sum_z\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).z'}|z'+e\rangle$$, proceeding from here the final step that I got was $$\dfrac{1}{\sqrt{2^n/|C_2|}}\sum_{z'+u\in C_2^{\perp}}(-1)^{(x+v)z'}|z'+e_2\rangle$$, Now is this correct, if so how do I proceed, and what should be the answer?